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combinations | Backtracking and recursion || c++ || easy Implementation | backtracking-and-recursion-c-easy-implem-se2i | Intuition\n Describe your first thoughts on how to solve this problem. \nIntution is basically picking number or notpicking that number.\n\n# Approach\n Describ | M_S_L | NORMAL | 2023-08-01T08:56:55.836443+00:00 | 2023-08-01T08:56:55.836478+00:00 | 13 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIntution is basically picking number or notpicking that number.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThis method uses a binary tree-like recursion structure where, at each step, it either includes the cu... | 5 | 0 | ['Backtracking', 'Recursion', 'C++'] | 0 |
combinations | Backtracking Solution | C++ | backtracking-solution-c-by-ayumsh-a5ov | Code\n\nclass Solution {\npublic:\n void backtrack(vector<vector<int>> &ans,vector<int> &temp,int ind, int n, int k)\n { \n if(k==0)\n {\n | ayumsh | NORMAL | 2023-08-01T07:07:39.482172+00:00 | 2023-08-01T07:08:12.816472+00:00 | 841 | false | # Code\n```\nclass Solution {\npublic:\n void backtrack(vector<vector<int>> &ans,vector<int> &temp,int ind, int n, int k)\n { \n if(k==0)\n {\n ans.push_back(temp);\n return;\n }\n for(int i=ind;i<=n-k+1;i++) \n {\n temp.push_back(i);\n ... | 5 | 0 | ['C++'] | 0 |
combinations | Beats 100% + Video | Java C++ Python | beats-100-video-java-c-python-by-jeevank-pet6 | \n\n\nclass Solution {\n List<List<Integer>> ans;\n public List<List<Integer>> combine(int n, int k) {\n ans = new ArrayList<>();\n helper(1 | jeevankumar159 | NORMAL | 2023-08-01T03:02:01.843938+00:00 | 2023-08-01T03:02:01.843968+00:00 | 1,227 | false | <iframe width="560" height="315" src="https://www.youtube.com/embed/DcUkm8CPeaY" title="YouTube video player" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" allowfullscreen></iframe>\n\n```\nclass Solution {\n List<List<Integer>> ans;\n ... | 5 | 1 | ['C', 'Python', 'Java'] | 0 |
combinations | Easy C++ Solution | easy-c-solution-by-2005115-2qmv | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nRecursive Backtracking:\n\nThe function combine calls the recursive funct | 2005115 | NORMAL | 2023-07-26T08:54:43.722144+00:00 | 2023-07-26T08:54:43.722164+00:00 | 532 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nRecursive Backtracking:\n\nThe function combine calls the recursive function solve to find all possible combinations of k numbers from the set [1, 2, ..., n].\nThe solve function uses backtracking to generate all possible su... | 5 | 0 | ['Backtracking', 'Recursion', 'C++'] | 0 |
combinations | Fast quick backtracking method | fast-quick-backtracking-method-by-haresh-axhm | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nBACK TRACKING\n\n# Comp | haresh412 | NORMAL | 2023-05-24T16:10:50.526119+00:00 | 2023-05-24T16:10:50.526153+00:00 | 816 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nBACK TRACKING\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(n!)\n- Space complexity:\n<!-- Add your space complexity he... | 5 | 0 | ['Backtracking', 'Depth-First Search', 'C++'] | 1 |
combinations | [C++] | 2 Different Approaches | Easy Explanation | Clean Code | c-2-different-approaches-easy-explanatio-dcjm | \n# Generate All Possible Subset\n Describe your approach to solving the problem. \nThe approach is to find out all the possible subset and take only those whic | _horiZon_OP | NORMAL | 2023-02-17T18:53:02.523035+00:00 | 2023-02-17T18:53:51.334412+00:00 | 2,217 | false | \n# Generate All Possible Subset\n<!-- Describe your approach to solving the problem. -->\nThe approach is to find out all the possible subset and take only those which have a size == k\n\nRecommended Question ( Subset ) : [ https://leetcode.com/problems/subsets/]()\n\n# Complexity\n- Time complexity : 2^N\n<!-- Add y... | 5 | 0 | ['Backtracking', 'C++'] | 1 |
combinations | 77. Combinations with step by step explanation | 77-combinations-with-step-by-step-explan-y3hr | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nThis solution uses the backtracking algorithm to generate all the combina | Marlen09 | NORMAL | 2023-02-14T04:22:01.672296+00:00 | 2023-02-14T04:22:01.672338+00:00 | 1,922 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThis solution uses the backtracking algorithm to generate all the combinations of k numbers from the range [1, n]. The function backtrack starts with a starting point first and an empty list curr which will store the current... | 5 | 0 | ['Python', 'Python3'] | 0 |
combinations | C++ Solution | c-solution-by-pranto1209-pdto | Approach\n Describe your approach to solving the problem. \n Backtracking\n\n# Code\n\nclass Solution {\npublic:\n vector<vector<int>> ans;\n int nl;\n | pranto1209 | NORMAL | 2023-01-14T17:57:30.573080+00:00 | 2023-03-14T08:15:20.335772+00:00 | 1,022 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\n Backtracking\n\n# Code\n```\nclass Solution {\npublic:\n vector<vector<int>> ans;\n int nl;\n\n void solve(vector<int> &a, int id, int k) {\n if(k == 0) {\n ans.push_back(a);\n return;\n }\n for(... | 5 | 0 | ['C++'] | 0 |
combinations | javascript backtracking | javascript-backtracking-by-egor_sh-p96j | \n/**\n * @param {number} n\n * @param {number} k\n * @return {number[][]}\n */\nvar combine = function (n, k) {\n let output = [];\n\n const findCombinat | egor_sh | NORMAL | 2022-11-15T19:56:40.170482+00:00 | 2022-11-15T19:56:40.170516+00:00 | 1,012 | false | ```\n/**\n * @param {number} n\n * @param {number} k\n * @return {number[][]}\n */\nvar combine = function (n, k) {\n let output = [];\n\n const findCombinations = (start, combination) => {\n if (combination.length === k) {\n output.push([...combination]);\n return;\n }\n\n ... | 5 | 0 | ['JavaScript'] | 0 |
combinations | Python3 | Backtracking | Intuitive and Easy to Understand | python3-backtracking-intuitive-and-easy-peibz | ```\nclass Solution:\n def combine(self, n: int, k: int) -> List[List[int]]:\n \n nums = list(range(1, n+1))\n results = []\n \n | ahbar | NORMAL | 2022-07-06T10:41:50.625796+00:00 | 2022-07-06T10:41:50.625842+00:00 | 1,073 | false | ```\nclass Solution:\n def combine(self, n: int, k: int) -> List[List[int]]:\n \n nums = list(range(1, n+1))\n results = []\n \n def backtrack(i, ans):\n if len(ans) == k:\n results.append(ans[:])\n return\n \n for num ... | 5 | 0 | ['Backtracking', 'Depth-First Search', 'Python', 'Python3'] | 2 |
combinations | C++ || Easy and Simple Code || Backtracking | c-easy-and-simple-code-backtracking-by-a-2ckl | \nclass Solution {\npublic:\n void backTrack(int n,int k,int it,vector<vector<int>> &ans,vector<int> &res){\n \n if(res.size()==k){\n | agrasthnaman | NORMAL | 2022-02-17T13:31:06.844299+00:00 | 2022-02-17T13:31:06.844377+00:00 | 482 | false | ```\nclass Solution {\npublic:\n void backTrack(int n,int k,int it,vector<vector<int>> &ans,vector<int> &res){\n \n if(res.size()==k){\n ans.push_back(res);\n return;\n }\n for(int i=it;i<=n;i++){\n res.push_back(i);\n backTrack(n,k,i+1,ans,res)... | 5 | 0 | ['Backtracking', 'C', 'C++'] | 0 |
combinations | combinations || c++ || backtracking | combinations-c-backtracking-by-sheetaljo-8rxv | TC=O(n*(2^n))\nSC=O(n) this can be a topic of discussion\n\nsimilar to the question subsets\nhttps://leetcode.com/problems/subsets/discuss/1654354/3-sol.-oror-c | sheetaljoshi | NORMAL | 2021-12-29T08:38:31.817074+00:00 | 2021-12-29T08:42:17.097920+00:00 | 527 | false | TC=O(n*(2^n))\nSC=O(n) this can be a topic of discussion\n\nsimilar to the question subsets\nhttps://leetcode.com/problems/subsets/discuss/1654354/3-sol.-oror-c%2B%2B-oror-subsets-oror-backtracking-oror-bit-manipulation-oror-iterative\n\n```\nclass Solution {\npublic:\n void generate(vector<int>& nums,int idx,int k,... | 5 | 0 | ['Backtracking', 'Recursion', 'C', 'C++'] | 0 |
combinations | General approach to backtracking problems in C++ | general-approach-to-backtracking-problem-9c0g | Below I present a general approach to many standard backtracking problems. A similar version exists for JAVA here but I thought it would be a good idea to do th | udaiwalrohit | NORMAL | 2021-06-28T17:46:38.539309+00:00 | 2021-06-28T17:46:38.539376+00:00 | 184 | false | Below I present a general approach to many standard backtracking problems. A similar version exists for JAVA [here](https://leetcode.com/problems/subsets/discuss/27281/A-general-approach-to-backtracking-questions-in-Java-(Subsets-Permutations-Combination-Sum-Palindrome-Partitioning)) but I thought it would be a good id... | 5 | 0 | [] | 0 |
combinations | Python One Liner using Combinations - 76ms | python-one-liner-using-combinations-76ms-praf | The itertools module provides a combinations() method which can be used in this question.\n\n\nfrom itertools import combinations \nclass Solution:\n def com | mb557x | NORMAL | 2020-07-14T10:48:43.348307+00:00 | 2020-07-14T10:48:43.348340+00:00 | 276 | false | The ```itertools``` module provides a ```combinations()``` method which can be used in this question.\n\n```\nfrom itertools import combinations \nclass Solution:\n def combine(self, n: int, k: int) -> List[List[int]]:\n return [i for i in list(combinations([i for i in range(1, n+1)], k))]\n```\n | 5 | 2 | ['Python3'] | 0 |
combinations | {{ Swift }} Iterative Solution | swift-iterative-solution-by-nraptis-kdo7 | \nclass Solution {\n func combine(_ n: Int, _ k: Int) -> [[Int]] {\n \n var res = [[Int]]()\n var data = [Int](repeating: 0, count: k)\n | nraptis | NORMAL | 2019-06-19T18:20:58.341235+00:00 | 2019-06-19T18:20:58.341275+00:00 | 257 | false | ```\nclass Solution {\n func combine(_ n: Int, _ k: Int) -> [[Int]] {\n \n var res = [[Int]]()\n var data = [Int](repeating: 0, count: k)\n var i = 0\n \n while data[0] <= (n - k + 1) {\n \n data[i] += 1\n if data[i] > n {\n i -= 1... | 5 | 0 | [] | 0 |
combinations | 5 lines python solution, very easy to understand | 5-lines-python-solution-very-easy-to-und-mp9q | ```\nclass Solution:\n def combine(self, n, k):\n if n == k:\n return [list(range(1,n+1))]\n if k == 1:\n return [ [i] fo | haefjdhfsf | NORMAL | 2018-10-11T07:28:53.677797+00:00 | 2018-10-24T06:31:45.322458+00:00 | 547 | false | ```\nclass Solution:\n def combine(self, n, k):\n if n == k:\n return [list(range(1,n+1))]\n if k == 1:\n return [ [i] for i in range(1,n+1) ]\n return self.combine(n-1,k) + [ j + [n] for j in self.combine(n-1,k-1) ] | 5 | 0 | [] | 1 |
combinations | Python recursive solution | python-recursive-solution-by-google-sjtd | class Solution:\n # @return a list of lists of integers\n # 9:14\n def __init__(self):\n self.output = []\n \n def com | google | NORMAL | 2015-03-09T02:22:58+00:00 | 2015-03-09T02:22:58+00:00 | 3,731 | false | class Solution:\n # @return a list of lists of integers\n # 9:14\n def __init__(self):\n self.output = []\n \n def combine(self, n, k, pos=0, temp=None):\n temp = temp or []\n \n if len(temp) == k:\n self.output.append(temp[:]... | 5 | 0 | ['Python'] | 3 |
combinations | Time Limit Exceeded... Why? | time-limit-exceeded-why-by-hammerking-onb9 | It works for n = 4, k = 2, but TLE for 20, 16. Anyone knows why it's inefficient?\n\n```class Solution(object):\n def combine(self, n, k):\n res = [] | hammerking | NORMAL | 2016-09-19T02:13:28.098000+00:00 | 2016-09-19T02:13:28.098000+00:00 | 1,416 | false | It works for n = 4, k = 2, but TLE for 20, 16. Anyone knows why it's inefficient?\n\n```class Solution(object):\n def combine(self, n, k):\n res = []\n self.combineHelper(n, k, res, [], 1)\n return res\n \n def combineHelper(self, n, k, res, temp, i):\n if len(temp) == k:\n... | 5 | 0 | [] | 5 |
combinations | Backtracking to Combine Numbers: Fast, Simple, and No Looking Back! | backtracking-to-combine-numbers-fast-sim-x7zp | Intuition\n Describe your first thoughts on how to solve this problem. \nThe first thought? It\u2019s pretty straightforward\u2014if we gotta pick k numbers out | satya78550 | NORMAL | 2024-10-18T17:07:14.367180+00:00 | 2024-10-18T17:07:14.367233+00:00 | 532 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe first thought? It\u2019s pretty straightforward\u2014if we gotta pick k numbers out of n, we\u2019re basically aiming to get all possible combinations of size k from 1 to n. Backtracking instantly pops into mind because it\u2019s perf... | 4 | 0 | ['Python3'] | 0 |
combinations | simple and easy C++ solution 😍❤️🔥 | simple-and-easy-c-solution-by-shishirrsi-llu6 | \n# if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n###### Let\'s Connect on LinkedIn: www.linkedin.com/in/shishirrsiam\n###### Let\'s Connect on Facebook | shishirRsiam | NORMAL | 2024-08-23T09:47:47.591135+00:00 | 2024-08-23T09:47:47.591198+00:00 | 1,118 | false | \n# if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n###### Let\'s Connect on LinkedIn: www.linkedin.com/in/shishirrsiam\n###### Let\'s Connect on Facebook: www.fb.com/shishirrsiam\n\n\n\n\n# Code\n```cpp []\nclass Solution {\npublic:\n int n, k;\n vector<int>path, nums;\n vector<vector<int>> ans;\n vo... | 4 | 0 | ['Backtracking', 'C++'] | 3 |
combinations | ✅💯🔥Simple Code📌🚀| 🔥✔️Easy to understand🎯 | 🎓🧠Beginner friendly🔥| O(n) Time Complexity💀💯 | simple-code-easy-to-understand-beginner-azal3 | Solution tuntun mosi ki photo ke baad hai. Scroll Down\n\n# Code\njava []\nclass Solution {\n public List<List<Integer>> combine(int n, int k){\n List | atishayj4in | NORMAL | 2024-08-21T19:32:22.593521+00:00 | 2024-08-21T19:32:22.593559+00:00 | 158 | false | Solution tuntun mosi ki photo ke baad hai. Scroll Down\n\n# Code\n```java []\nclass Solution {\n public List<List<Integer>> combine(int n, int k){\n ... | 4 | 0 | ['Backtracking', 'C', 'Python', 'C++', 'Java', 'JavaScript'] | 0 |
combinations | ✅Easy and simple solution ✅clean code | easy-and-simple-solution-clean-code-by-a-hmkt | Guy\'s if you find this solution helpful \uD83D\uDE0A, PLEASE do UPVOTE. By doing that it motivate\'s me to create more better post like this \u270D\uFE0F\nFoll | ayushluthra62 | NORMAL | 2024-07-23T13:19:00.342817+00:00 | 2024-07-23T13:19:00.342864+00:00 | 470 | false | ***Guy\'s if you find this solution helpful \uD83D\uDE0A, PLEASE do UPVOTE. By doing that it motivate\'s me to create more better post like this \u270D\uFE0F***<br>\n**Follow me on LinkeDin [[click Here](https://www.linkedin.com/in/ayushluthra62/)]**\n\n# Complexity\n- Time complexity:\nO(NcK)\n\n# Code\n```\nclass Sol... | 4 | 0 | ['Backtracking', 'C++'] | 0 |
combinations | Solution By Dare2Solve | Detailed Explanation | Clean Code | solution-by-dare2solve-detailed-explanat-rzag | Explanation []\nauthorslog.com/blog/EtIaoat6mn\n\n# Code\n\ncpp []\nclass Solution {\npublic:\n vector<vector<int>> combine(int n, int k) {\n vector<v | Dare2Solve | NORMAL | 2024-07-20T17:38:43.954833+00:00 | 2024-07-20T17:38:43.954858+00:00 | 840 | false | ```Explanation []\nauthorslog.com/blog/EtIaoat6mn\n```\n# Code\n\n```cpp []\nclass Solution {\npublic:\n vector<vector<int>> combine(int n, int k) {\n vector<vector<int>> result;\n vector<int> combination;\n\n backtrack(n, k, 1, combination, result);\n return result;\n }\n\nprivate:\n ... | 4 | 0 | ['Backtracking', 'Python', 'C++', 'Java', 'Python3', 'JavaScript'] | 1 |
combinations | C++ and Python3 || Backtracking || Simple and Optimal | c-and-python3-backtracking-simple-and-op-0gx7 | \n# Code\nC++ []\nclass Solution {\npublic:\n void bt(int idx, int k, int n, vector<int> &comb, vector<vector<int>> &ans) {\n if(comb.size() == k) {\n | meurudesu | NORMAL | 2024-02-28T11:01:25.962614+00:00 | 2024-02-28T11:01:25.962636+00:00 | 598 | false | > \n# Code\n```C++ []\nclass Solution {\npublic:\n void bt(int idx, int k, int n, vector<int> &comb, vector<vector<int>> &ans) {\n if(comb.size() == k) {\n ans.push_back(comb);\n ... | 4 | 0 | ['Backtracking', 'Depth-First Search', 'Recursion', 'C++', 'Python3'] | 2 |
combinations | Easy Beginner friendly Solution || Java || C++ || Python | easy-beginner-friendly-solution-java-c-p-1b8w | Intuition\n Describe your first thoughts on how to solve this problem. \nThis Java code provides a solution to generate combinations of k elements from the numb | Aditya_Mohan_Gupta02 | NORMAL | 2024-02-13T19:02:39.587008+00:00 | 2024-02-13T19:02:39.587038+00:00 | 319 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThis Java code provides a solution to generate combinations of k elements from the numbers 1 to n.\nIn summary, this code employs a recursive backtracking approach to generate all possible combinations of k elements from the numbers 1 to ... | 4 | 0 | ['Java'] | 0 |
combinations | Best Solution | best-solution-by-kumar21ayush03-lc8n | Approach\nRecursion\n\n# Complexity\n- Time complexity:\nO(2^n)\n\n- Space complexity:\nO(n) \n\n# Code\n\nclass Solution {\npublic:\n void solve(int i, int | kumar21ayush03 | NORMAL | 2023-08-02T05:37:00.391601+00:00 | 2023-08-02T05:37:00.391629+00:00 | 19 | false | # Approach\nRecursion\n\n# Complexity\n- Time complexity:\n$$O(2^n)$$\n\n- Space complexity:\n$$O(n)$$ \n\n# Code\n```\nclass Solution {\npublic:\n void solve(int i, int k, vector<int>& temp, vector<vector<int>>& ans) {\n if (k == 0) {\n ans.push_back(temp);\n return;\n }\n ... | 4 | 0 | ['C++'] | 0 |
combinations | Simple and basic approach using recursion of take and nottake case | simple-and-basic-approach-using-recursio-8pdr | Intuition\n simply used take and nottake cases for combinations \n\n# Approach\n used backtracking first push element into array untill size of temp bacomes K a | mahadev_lover | NORMAL | 2023-08-01T08:22:04.272204+00:00 | 2023-08-01T08:22:04.272239+00:00 | 92 | false | # Intuition\n<!-- simply used take and nottake cases for combinations -->\n\n# Approach\n<!-- used backtracking first push element into array untill size of temp bacomes K and then pop the element that is pushed so that next element can be taken -->\n\n# Complexity\n- Time complexity:\n<!-- O(2^k) -->\n\n- Space comple... | 4 | 0 | ['Array', 'Backtracking', 'C++'] | 0 |
combinations | Bitmasking || Easily explained in Hindi || Clean and concise code | bitmasking-easily-explained-in-hindi-cle-4kfd | Approach\n1. We have to choose exactly K numbers out of N numbers.\n2. To choose that we will create bit masks.\n3. What bit masks will do is, we have all the n | ashay028 | NORMAL | 2023-08-01T07:49:47.657507+00:00 | 2023-08-01T07:49:47.657531+00:00 | 57 | false | # Approach\n1. We have to choose exactly K numbers out of N numbers.\n2. To choose that we will create bit masks.\n3. What bit masks will do is, we have all the numbers which have only K set bits and it is less than 2^N (i.e. total number of bits are less equal to N) so these will help to choose the K numbers. \n\nSo w... | 4 | 0 | ['C++'] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | [Python3] Easy Python3 BFS + Graph Like | python3-easy-python3-bfs-graph-like-by-l-kla5 | nextLetter is like a graph. a -> bc, b -> ac, c -> ab. \nSince we are doing BFS with the next node in the graph already sorted, \nwe are guaranteed to have a so | localhostghost | NORMAL | 2020-04-18T21:55:14.264218+00:00 | 2020-04-20T11:53:01.555206+00:00 | 5,715 | false | ` nextLetter ` is like a graph. `a -> bc`, `b -> ac`, `c -> ab`. \nSince we are doing BFS with the next node in the graph already sorted, \nwe are guaranteed to have a sorted queue with all the element of the \nsame length.\n\nBreak when the first string in the queue is length `n` and find the `k`th value in the queue.... | 130 | 1 | [] | 12 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | [C++/Java] DFS and Math | cjava-dfs-and-math-by-votrubac-l5d3 | Aproach 1: DFS\nHere, DFS builds all possible strings in a lexicographical order. Each time a string is built, we decrease the global counter k. When k reaches | votrubac | NORMAL | 2020-04-18T16:03:12.591819+00:00 | 2020-05-14T06:26:07.636302+00:00 | 11,352 | false | #### Aproach 1: DFS\nHere, DFS builds all possible strings in a lexicographical order. Each time a string is built, we decrease the global counter `k`. When `k` reaches zero, we short-circut DFS and build the resulting string right-to-left. \n\n> Exploring references and optional parameters for a concise code.\n\n```cp... | 110 | 3 | [] | 30 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | 🚀 Beats 100% | Algorithm: Depth-First Search (DFS) for K-th Happy String | Explaination with Video | beats-100-algorithm-depth-first-search-d-kjjr | Youtube🚀🚀 Beats 100% | Algorithm: Depth-First Search (DFS) for K-th Happy String | Explanation with Video 🔍🔼 Please Upvote🔼 Please Upvote🔼 Please Upvote🔼 Please | rahulvijayan2291 | NORMAL | 2025-02-19T04:58:50.055162+00:00 | 2025-02-19T04:58:50.055162+00:00 | 15,454 | false | # Youtube
https://youtu.be/fsFdfuF_8zs
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# 🚀 **🚀 Beats 100% | Algorithm: Depth-First Search (DFS) for K-th Happy String | Explanation with Video 🔍**
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# **🔼 Please Upvote**
# **🔼 Please Upvo... | 78 | 1 | ['String', 'Depth-First Search', 'C', 'C++', 'Java', 'Python3', 'JavaScript'] | 7 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | C++ Recursive fully explained solution | c-recursive-fully-explained-solution-by-1wt3a | \n//This is my first try to put code into discussion part hope you like it and if like please upvote :)\n\nclass Solution {\npublic:\n \n void solve(vecto | harsh_114 | NORMAL | 2020-04-19T03:49:10.235671+00:00 | 2020-04-19T03:50:39.490401+00:00 | 3,099 | false | ```\n//This is my first try to put code into discussion part hope you like it and if like please upvote :)\n\nclass Solution {\npublic:\n \n void solve(vector<string>& ans,string& cur,int n,int& k,vector<char>& v){\n //before base cases please refer lower part :)\n if(ans.size()==k){\n //... | 53 | 0 | ['Recursion', 'C'] | 5 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Easiest solution using bfs---C++ | easiest-solution-using-bfs-c-by-mr_233-juzm | \nclass Solution {\npublic:\n string getHappyString(int n, int k) {\n vector<string> ans;\n queue<string> q;\n q.push("a");\n q.p | mr_233 | NORMAL | 2020-04-18T16:01:48.680325+00:00 | 2020-04-19T07:20:29.988701+00:00 | 3,023 | false | ```\nclass Solution {\npublic:\n string getHappyString(int n, int k) {\n vector<string> ans;\n queue<string> q;\n q.push("a");\n q.push("b");\n q.push("c");\n while(!q.empty())\n {\n string x=q.front();\n q.pop();\n if(x.length()==n)\n... | 43 | 0 | [] | 8 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Combinatorics+ binary expression||C++ Py3 beats 100% | combinatorics-binary-expressionc-beats-1-a6k7 | IntuitionThere are exactly3×2n−1happy strings of lengthn(combinatorical question).The set of happy strings can be divided into 3 subsets consisting of strings b | anwendeng | NORMAL | 2025-02-19T00:21:49.052019+00:00 | 2025-02-19T04:19:51.096889+00:00 | 5,627 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
There are exactly $3\times 2^{n-1}$ happy strings of length $n$ (combinatorical question).
The set of happy strings can be divided into 3 subsets consisting of strings beginning with a, b & c.
The kth one must be in 1 of them. So the expre... | 42 | 0 | ['String', 'Bit Manipulation', 'Combinatorics', 'Bitmask', 'C++', 'Python3'] | 11 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Java - Short and understandable #NoOverdoing | java-short-and-understandable-nooverdoin-ln84 | I am relatively new to this, so trust me if I am able to figure this one out, you can too :)!\n\nI basically generate all permutations with s[i] != s[i+1] using | narihanellaithy | NORMAL | 2020-04-18T16:22:14.453706+00:00 | 2020-04-18T16:22:37.777005+00:00 | 2,697 | false | **I am relatively new to this, so trust me if I am able to figure this one out, you can too :)!**\n\nI basically generate all permutations with s[i] != s[i+1] using this check as I append my next permutation\n```\nres.charAt(res.length()-1) != arr[i]\n```\n\n\n```\nclass Solution {\n public String getHappyString(int... | 24 | 2 | [] | 8 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | 😊 Easy Solution with BFS | Generate Happy Strings! | easy-solution-with-bfs-generate-happy-st-rnne | Intuition 💡Ahappy stringis a string that:✅ Consists only of the letters'a', 'b', 'c'.✅ No two adjacent characters are the same.Given two integersnandk, we need | lasheenwael9 | NORMAL | 2025-02-19T06:47:13.724321+00:00 | 2025-02-19T06:47:13.724321+00:00 | 2,477 | false | # Intuition 💡
A **happy string** is a string that:
✅ Consists only of the letters **'a', 'b', 'c'**.
✅ No two adjacent characters are the same.
Given two integers **n** and **k**, we need to generate a list of all **happy strings** of length **n** in **lexicographical order**, then return the **kth** string. ... | 23 | 0 | ['String', 'Backtracking', 'Breadth-First Search', 'Queue', 'Python', 'C++'] | 1 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | [Python] 7-line DFS with yield | python-7-line-dfs-with-yield-by-yanruche-9o82 | Idea\nThe property of DFS helps us iterate through all the possibilities in ascending order automatically. All we need to do is to output the kth result.\n\nPyt | yanrucheng | NORMAL | 2020-04-18T16:01:16.066999+00:00 | 2020-04-18T16:01:16.067047+00:00 | 1,820 | false | **Idea**\nThe property of DFS helps us iterate through all the possibilities in ascending order automatically. All we need to do is to output the kth result.\n\n**Python**\n```\nclass Solution:\n def getHappyString(self, n: int, k: int) -> str:\n def generate(prev):\n if len(prev) == n:\n ... | 23 | 3 | [] | 5 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | ✅✅🔥BEATS 99.09% SIMPLEST PROGRAM | JAVA 🔥 Python 🔥 C++ & C 🔥 | Simplest Program | O(2^n) & O(n) | beats-9909-simplest-program-java-python-d8toi | IntuitionGoal is to generates the k-th lexicographically smallest "happy string" of length n. A "happy string" is defined as a string that:
Consists only of the | peEAWW4Y33 | NORMAL | 2025-02-19T16:38:52.580213+00:00 | 2025-02-19T16:38:52.580213+00:00 | 1,291 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Goal is to generates the k-th lexicographically smallest "happy string" of length n. A "happy string" is defined as a string that:
- Consists only of the characters a, b, and c
- Has no two adjacent characters that are the same.
# A... | 21 | 0 | ['String', 'Backtracking', 'C', 'Python', 'C++', 'Java'] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Most Optimal Solution ✅ C++| Java | Python | JavaScript | most-optimal-solution-c-java-python-java-aqcn | ⬆️Upvote if it helps ⬆️Connect with me on Linkedin [Bijoy Sing]Solution in C++, Python, Java, and JavaScriptHere’s a detailed explanation of the solution:Intuit | BijoySingh7 | NORMAL | 2025-02-19T07:08:37.802730+00:00 | 2025-02-19T07:08:37.802730+00:00 | 2,082 | false | # ⬆️Upvote if it helps ⬆️
---
## Connect with me on Linkedin [Bijoy Sing]
---
###### *Solution in C++, Python, Java, and JavaScript*
```cpp []
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
class Solution {
public:
string getHappyString(int n, int k) {
queue<string> q;
... | 20 | 1 | ['String', 'Backtracking', 'Queue', 'Heap (Priority Queue)', 'Python', 'C++', 'Java', 'Python3', 'JavaScript'] | 2 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Python3 O(n) solution using math with clear explanation | python3-on-solution-using-math-with-clea-tdix | let\'s consider n=3 and k=9\nThe lexicographical order of the strings are ["aba", "abc", "aca", "acb", "bab", "bac", "bca", "bcb", "cab", "cac", "cba", "cbc"]\n | dark_wolf_jss | NORMAL | 2020-06-08T03:23:38.758499+00:00 | 2020-06-08T03:23:38.758534+00:00 | 1,612 | false | let\'s consider n=3 and k=9\nThe lexicographical order of the strings are ["aba", "abc", "aca", "acb", "bab", "bac", "bca", "bcb", "cab", "cac", "cba", "cbc"]\n\nWe can observe that each element a,b,c has repeated 2**(n-1) times as the first character which is 4 in our case they are ["aba", "abc", "aca", "acb"], ["bab... | 19 | 0 | ['Math', 'Python3'] | 2 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Java O(n) solution | java-on-solution-by-chenwei713-zbmz | \nclass Solution {\n public String getHappyString(int n, int k) {\n if(n == 0) {\n return "";\n }\n \n // K is too lar | chenwei713 | NORMAL | 2020-04-18T16:39:19.692023+00:00 | 2020-04-18T16:39:19.692156+00:00 | 1,608 | false | ```\nclass Solution {\n public String getHappyString(int n, int k) {\n if(n == 0) {\n return "";\n }\n \n // K is too large\n if(k > (int)3 * Math.pow(2, n - 1)) {\n return "";\n }\n \n // Make K start from 0\n k--;\n \n ... | 16 | 3 | [] | 5 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | 🔥BEATS 💯 % 🎯 |✨SUPER EASY BEGINNERS 👏| JAVA | C | C++ | PYTHON| JAVASCRIPT | DART | beats-super-easy-beginners-java-c-c-pyth-a90w | IntuitionThe problem requires generating lexicographically sorted "happy strings" of lengthnusing the characters'a','b', and'c'. Ahappy stringis defined as a st | CodeWithSparsh | NORMAL | 2025-02-19T08:42:23.529110+00:00 | 2025-02-19T08:42:23.529110+00:00 | 1,314 | false | 
---
# **Intuition**
The problem requires generating lexicographically sorted "happy strings" of length **n** using the characters `'a'`, `'b'`, and `'c'`. A **happy string** is defined as a string whe... | 15 | 0 | ['String', 'Backtracking', 'C', 'C++', 'Java', 'Python3', 'JavaScript', 'Dart'] | 3 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Video solution | Intuition and Approach explained in detail | C++ | Backtracking | video-solution-intuition-and-approach-ex-frtt | Hello every one i have created a video solution for this problem and solved this video by developing intutition for backtracking (its in hindi), i am pretty sur | _code_concepts_ | NORMAL | 2024-04-30T05:38:52.795287+00:00 | 2024-04-30T05:39:42.974978+00:00 | 587 | false | Hello every one i have created a video solution for this problem and solved this video by developing intutition for backtracking (its in hindi), i am pretty sure you will never have to look for solution for this video ever again after watching this.\n\n\nThis video is the part of my playlist "Master backtracking".\nVid... | 15 | 0 | ['C++'] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | ✅✅100% Beats || Simple Backtracking Approach | 100-beats-simple-backtracking-approach-b-cdoj | IntuitionA "happy string" of lengthnconsists of charactersa,b, andcsuch that no two adjacent characters are the same. To find thek-th lexicographically smallest | arunk_leetcode | NORMAL | 2025-02-19T04:15:46.378656+00:00 | 2025-02-19T04:15:46.378656+00:00 | 5,217 | false | # Intuition
A "happy string" of length `n` consists of characters `a`, `b`, and `c` such that no two adjacent characters are the same. To find the `k`-th lexicographically smallest happy string, we can generate all valid strings in lexicographical order and return the `k`-th one.
# Approach
1. Use **recursive backtrac... | 14 | 0 | ['String', 'Backtracking', 'Python', 'C++', 'Java', 'Python3'] | 5 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Simple Java solution::Backtracking | simple-java-solutionbacktracking-by-2018-jwga | DO UPVOTE INCASE YOU FIND IT HELPFUL!\n```\nclass Solution {\n List list=new ArrayList<>();\n StringBuilder sb = new StringBuilder();\n public String g | 20184152 | NORMAL | 2021-06-05T02:12:47.713459+00:00 | 2021-06-05T02:12:47.713489+00:00 | 1,462 | false | DO UPVOTE INCASE YOU FIND IT HELPFUL!\n```\nclass Solution {\n List<String> list=new ArrayList<>();\n StringBuilder sb = new StringBuilder();\n public String getHappyString(int n, int k) {\n backtrack(n,\' \');\n Collections.sort(list);\n if(list.size()<k) return "";\n else return l... | 14 | 1 | ['Backtracking', 'Java'] | 5 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | backtracking||beats 99.64%|| | backtrackingbeats-9964-by-srujitha_vutla-0xyx | \nthis question is similar to https://leetcode.com/problems/letter-tile-possibilities/\n 1.one of the possible way to solve this question is generating all stri | srujitha_vutla | NORMAL | 2024-02-18T04:39:26.342639+00:00 | 2024-02-18T04:39:26.342657+00:00 | 328 | false | \nthis question is similar to https://leetcode.com/problems/letter-tile-possibilities/\n 1.one of the possible way to solve this question is generating all strings of length n and returning kth string\n 2.here ... | 13 | 1 | [] | 1 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Python 3 || 9 lines, binary map w/ example || T/S: 98% / 99% | python-3-9-lines-binary-map-w-example-ts-st2m | https://leetcode.com/problems/the-k-th-lexicographical-string-of-all-happy-strings-of-length-n/submissions/937031498/updated run 2/18/2025I could be wrong, but | Spaulding_ | NORMAL | 2023-04-20T17:54:08.220366+00:00 | 2025-02-19T00:23:42.400658+00:00 | 629 | false | ```
class Solution:
def getHappyString(self, n: int, k: int) -> str:
d = {'a':'bc','b':'ac','c':'ab'} # Example: n = 5, k = 20
div,r = divmod(k-1,2**(n-1)) # div,r = divmod(19,16) = 1,3
if div > 2: return ''
prev = ans = 'abc'[div] ... | 13 | 0 | ['Python3'] | 1 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | [C++] Straightforward DFS. Skip appending same char. | c-straightforward-dfs-skip-appending-sam-n0kk | \nUPDATE:\n\nAfter contest, I think we can avoid storing all the strings in the vector, but only keep the pontential result string(cur).\nScroll down to check f | xzj104 | NORMAL | 2020-04-18T16:01:08.120985+00:00 | 2020-04-19T02:25:45.824417+00:00 | 1,491 | false | \n***UPDATE:***\n\nAfter contest, I think we can avoid storing all the strings in the vector, but only keep the pontential result string(cur).\nScroll down to check first solution.\n```\nclass Solution {\npublic:\n string getHappyString(int n, int k) {\n string cur = "";\n left = k;\n dfs(cur, n... | 13 | 1 | [] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | [Java] Concise and clean solution, imagine problem as a tree, 0ms. | java-concise-and-clean-solution-imagine-24li0 | Update\nChanged string concatenation to StringBuilder. Performance gain from 11ms to 0ms.\n\n### Solution\n\nMy idea was to imagine the problem as a tree:\n1. O | hydro_0 | NORMAL | 2020-04-19T08:59:18.643865+00:00 | 2020-04-20T08:49:17.911962+00:00 | 1,184 | false | ### Update\nChanged string concatenation to `StringBuilder`. Performance gain from `11ms` to `0ms`.\n\n### Solution\n\nMy idea was to imagine the problem as a tree:\n1. On the first step we have three branches: `a`, `b` and `c`.\n2. On each next step we have two branches: if node is `a`: `b` and `c`, for `b`: `a` and `... | 11 | 0 | ['Tree', 'Java'] | 5 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | 0 ms. O(n). No branching. Combinatorics. Beats 100.00% | 0-ms-no-branching-combinatorics-beats-10-r6xm | IntuitionA beautiful problem with a beautiful solution. The first letter is chosen from threea,borc. The second and subsequent letters are chosen from the two r | nodeforce | NORMAL | 2025-02-19T09:26:40.815954+00:00 | 2025-02-19T15:23:12.097402+00:00 | 905 | false | # Intuition
A beautiful problem with a beautiful solution. The first letter is chosen from three `a`, `b` or `c`. The second and subsequent letters are chosen from the two remaining. If we denote it as `0` or `1`. Which brings us to the binary code of the number.
# Approach
First, let's calculate the number of happy s... | 10 | 0 | ['String', 'C', 'Combinatorics', 'Bitmask', 'Python', 'C++', 'Java', 'TypeScript', 'Python3', 'JavaScript'] | 1 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Python | Simple Backtracking | O(2^n), O(n) | Beats 100% | python-simple-backtracking-on-2n-on-beat-jflm | CodeComplexity
Time complexity:O(2n). backtrack() tries to make every happy string, 3∗2n−1total since 3 characters in first index and 2 for remaining, simplifi | 2pillows | NORMAL | 2025-02-19T01:01:34.455611+00:00 | 2025-02-19T04:39:29.875289+00:00 | 1,229 | false | # Code
```python3 []
class Solution:
def getHappyString(self, n: int, k: int) -> str:
count = 0 # tracks number of completed strings made
def backtrack(i, chars):
nonlocal count
if i == n:
count += 1 # finished a string
return "".join(chars)... | 10 | 1 | ['String', 'Backtracking', 'Python3'] | 2 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | C++ BFS Without using sort. | c-bfs-without-using-sort-by-abhinandanka-dinw | \n string getHappyString(int n, int k) {\n vector<string> all;\n queue<string> q;\n q.push("a");\n q.push("b");\n q.push(" | abhinandankainth1999 | NORMAL | 2020-04-18T20:22:41.690075+00:00 | 2020-04-18T20:22:41.690119+00:00 | 504 | false | ```\n string getHappyString(int n, int k) {\n vector<string> all;\n queue<string> q;\n q.push("a");\n q.push("b");\n q.push("c");\n int t = n;\n while(t--){\n int sz = q.size();\n while(sz--){\n string s = q.front();\n ... | 10 | 0 | [] | 3 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | 🚀 Beats 100% | Simple & Efficient O(n) Solution | No Backtracking Needed! 🚀 | beats-100-simple-efficient-on-solution-n-28j4 | IntuitionThe intuition behind this code is to efficiently determine thek-th lexicographically smallest happy stringwithout generating all possible strings.
Und | aayush8910sh | NORMAL | 2025-02-19T05:55:34.814464+00:00 | 2025-02-19T05:55:34.814464+00:00 | 402 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The intuition behind this code is to efficiently determine the **k-th lexicographically smallest happy string** without generating all possible strings.
1. **Understanding the Pattern**:
- A happy string of length `n` follows a struct... | 9 | 0 | ['Math', 'Bit Manipulation', 'C++'] | 3 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Easy to understand cpp solution | easy-to-understand-cpp-solution-by-vp-krf8 | we will check if our previous character is not the same as the current one. If not, we will push the current character. This is the same condition as the happy | Vp- | NORMAL | 2021-07-22T12:36:00.155991+00:00 | 2021-07-22T12:36:00.156052+00:00 | 540 | false | we will check if our previous character is not the same as the current one. If not, we will push the current character. This is the same condition as the happy string.\n\n```cpp\nclass Solution {\npublic:\n void helper(int n,string check,vector<string>& ans)\n {\n if(n == 0){\n ans.push_back(che... | 9 | 0 | ['C++'] | 2 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Easy Approach || Recursion | easy-approach-recursion-by-muhammednihas-fbqe | IntuitionThe problem requires us to generate lexicographically sorted "happy strings" of lengthnand return thek-th one. A happy string is defined as a string co | muhammednihas2218 | NORMAL | 2025-02-19T04:37:19.326680+00:00 | 2025-02-21T05:09:36.998083+00:00 | 795 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem requires us to generate lexicographically sorted "happy strings" of length `n` and return the `k`-th one. A happy string is defined as a string containing only the characters `a`, `b`, and `c` with no two adjacent characters be... | 8 | 1 | ['String', 'Backtracking', 'Recursion', 'C', 'C++', 'Java', 'Go', 'Python3', 'JavaScript'] | 1 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | ✔️ Easiest Solution & Beginner Friendly ✔️ | 🔥 Recursion + Backtracking 🔥 | 2 Solutions | easiest-solution-beginner-friendly-recur-nen0 | IntuitionImagine you are forming a word using only the lettersa, b, cbut with one rule—no twoconsecutive letters can be the same. You need to generate all possi | ntrcxst | NORMAL | 2025-02-19T03:57:50.372078+00:00 | 2025-02-19T03:57:50.372078+00:00 | 971 | false | # Intuition
Imagine you are forming a word using only the letters `a, b, c` but with one rule—no two **consecutive letters can be the same**. You need to generate all possible **valid words of a given length** `n`, sort them **lexicographically**, and find the `k-th` one. Instead of **brute force**, we use **backtracki... | 8 | 0 | ['Array', 'String', 'Binary Search', 'Backtracking', 'Greedy', 'Tree', 'Depth-First Search', 'Breadth-First Search', 'C++', 'Java'] | 1 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | The k-th Lexicographical String of All Happy Strings of Length n || simple backtracking || O(3^n) | the-k-th-lexicographical-string-of-all-h-jl4v | IntuitionThe problem requires us to generate all possible "happy strings" of length n where the string consists of the characters 'a', 'b', and 'c', with no two | Mirin_Mano_M | NORMAL | 2025-02-19T08:25:26.918502+00:00 | 2025-02-19T08:25:58.845239+00:00 | 95 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem requires us to generate all possible "happy strings" of length n where the string consists of the characters 'a', 'b', and 'c', with no two consecutive characters being the same. After generating all the happy strings, we need t... | 6 | 0 | ['String', 'Backtracking', 'C++', 'Java'] | 1 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | ✔️ Easiest Solution & Beginner Friendly ✔️ | 🔥 Recursion + Backtracking 🔥 | 2 Solutions | easiest-solution-beginner-friendly-recur-i08p | IntuitionImagine you are forming a word using only the lettersa, b, cbut with one rule—no twoconsecutive letters can be the same. You need to generate all possi | ntrcxst | NORMAL | 2025-02-19T03:57:45.588724+00:00 | 2025-02-19T03:57:45.588724+00:00 | 330 | false | # Intuition
Imagine you are forming a word using only the letters `a, b, c` but with one rule—no two **consecutive letters can be the same**. You need to generate all possible **valid words of a given length** `n`, sort them **lexicographically**, and find the `k-th` one. Instead of **brute force**, we use **backtracki... | 6 | 0 | ['Array', 'String', 'Binary Search', 'Backtracking', 'Greedy', 'Tree', 'Depth-First Search', 'Breadth-First Search', 'C++', 'Java'] | 2 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Recursion | Backtracking | Solution for LeetCode#1415 | recursion-backtracking-solution-for-leet-zuo5 | IntuitionTo solve the problem of generating the k-th lexicographical happy string of length n, the first thought is to generate all possible happy strings of le | samir023041 | NORMAL | 2025-02-19T02:10:11.843965+00:00 | 2025-02-19T02:10:11.843965+00:00 | 1,462 | false | 
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
To solve the problem of generating the k-th lexicographical happy string of length n, the first thought is to generate all p... | 6 | 0 | ['Backtracking', 'Recursion', 'Java'] | 1 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | [Working Explained Using Permutation Concept | Complete Walkthrough] | working-explained-using-permutation-conc-wv8h | \n/*\n 3 Key Important things to know to solve any Backtracking problem\n ----------------------------------------------------------------\n\n 1.) Choi | _LearnToCode_ | NORMAL | 2022-09-27T11:48:26.916361+00:00 | 2022-09-27T11:51:40.981173+00:00 | 495 | false | ```\n/*\n 3 Key Important things to know to solve any Backtracking problem\n ----------------------------------------------------------------\n\n 1.) Choices [Search Space from which we\'ve to choose items]\n ==========================================================\n search_space = {\'a\', \'b\', \... | 6 | 0 | ['Java'] | 3 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Backtracking solution C++ | backtracking-solution-c-by-akansha_22-96w0 | ```\nclass Solution {\npublic:\n \n vector v={\'a\',\'b\',\'c\'};\n \n void helper(vector &ans,string t,int n,char prev)\n {\n if(n==0)\n | akansha_22 | NORMAL | 2021-05-16T11:35:20.671433+00:00 | 2021-05-16T11:35:20.671473+00:00 | 553 | false | ```\nclass Solution {\npublic:\n \n vector<char> v={\'a\',\'b\',\'c\'};\n \n void helper(vector<string> &ans,string t,int n,char prev)\n {\n if(n==0)\n {\n ans.push_back(t);\n return ;\n }\n \n for(int i=0;i<v.size();i++)\n {\n if... | 6 | 0 | ['Backtracking', 'C'] | 2 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Best Approach to Practice Backtracking. | best-approach-to-practice-backtracking-b-f9bd | NoteMy dear Friend here I am Not going for Optimize or Time Complexity Based Approach, i know we can solve it using DFS and Queue Optimally.My only goal is to p | Tanmay_bhure | NORMAL | 2025-02-19T19:28:59.663699+00:00 | 2025-02-19T19:28:59.663699+00:00 | 103 | false | # Note
***My dear Friend here I am Not going for Optimize or Time Complexity Based Approach, i know we can solve it using DFS and Queue Optimally.
My only goal is to provide best backtracking approach So that You can Practice It, and get view of Backtracking Working.***
# Intuition
The problem asks to generate all "h... | 5 | 0 | ['String', 'Backtracking', 'C++', 'Java', 'Python3'] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | ✅ One Line Solution | one-line-solution-by-mikposp-hml7 | (Disclaimer: this is not an example to follow in a real project - it is written for fun and training mostly)Code #1.1 - One LineTime complexity:O(3n). Space com | MikPosp | NORMAL | 2025-02-19T09:49:05.459028+00:00 | 2025-02-19T09:50:18.612535+00:00 | 593 | false | (Disclaimer: this is not an example to follow in a real project - it is written for fun and training mostly)
# Code #1.1 - One Line
Time complexity: $$O(3^n)$$. Space complexity: $$O(n)$$.
```python3
class Solution:
def getHappyString(self, n: int, k: int) -> str:
return next(islice((s for p in product(*['... | 5 | 0 | ['String', 'Combinatorics', 'Python', 'Python3'] | 2 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Easy Peasy, Daily Grind! 🌩️ | easy-peasy-daily-grind-by-_abhiramks-i42y | IntuitionApproachComplexity
Time complexity: O(2n)
Space complexity: O(2n)
Code | _abhiramks | NORMAL | 2025-02-19T04:24:01.867941+00:00 | 2025-02-19T04:24:01.867941+00:00 | 15 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: O(2n)
- Space complexity: O(2n)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```dart []
🔥 steps
<!-- 1. Create an empt... | 5 | 0 | ['Array', 'String', 'Backtracking', 'Recursion', 'Dart'] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Beats 100% | Easy Explaination | Combinatorics & Maths 👍 | beats-100-easy-explaination-combinatoric-lb25 | IntuitionThe problem requires generating the k-th lexicographically smallest happy string of lengthn. A happy string is defined as a string consisting only of c | com_gaurav_dev | NORMAL | 2025-02-19T04:12:45.335750+00:00 | 2025-02-19T04:12:45.335750+00:00 | 369 | false | # Intuition
The problem requires generating the k-th lexicographically smallest happy string of length `n`. A happy string is defined as a string consisting only of characters `a`, `b`, and `c` such that no two adjacent characters are the same.
The key observation here is that for each position in the string, we can c... | 5 | 0 | ['Bit Manipulation', 'Combinatorics', 'C++'] | 2 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Fast and Simple C++ | fast-and-simple-c-by-arpit_sat-qv6h | \nvoid find_all(int n,string &curr,vector<string> &v)\n{\n\tif(curr.size()==n)\n\t{\n\t\tv.push_back(curr);\n\t\treturn ;\n\t}\n\tfor(char i=\'a\';i<=\'c\';i++) | Arpit_Sat | NORMAL | 2021-03-25T14:50:25.921600+00:00 | 2021-03-25T14:50:25.921649+00:00 | 285 | false | ```\nvoid find_all(int n,string &curr,vector<string> &v)\n{\n\tif(curr.size()==n)\n\t{\n\t\tv.push_back(curr);\n\t\treturn ;\n\t}\n\tfor(char i=\'a\';i<=\'c\';i++)\n\t{\n\t\tif(curr.size()==0||i!=curr.back())\n\t\t{\n\t\t\tcurr.push_back(i);\n\t\t\tfind_all(n,curr,v);\n\t\t\tcurr.pop_back();\n\t\t}\n\t}\n}\nstring getH... | 5 | 0 | [] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Python, O(n) (faster than 99%, memory usage less than 100%) | python-on-faster-than-99-memory-usage-le-o5of | We calculate how many possible happy substrings of given length minus one exist \u2013 using this number, k, a list of possible letters and the last used letter | karbayev | NORMAL | 2020-04-20T22:38:20.223923+00:00 | 2020-05-09T12:37:55.527055+00:00 | 675 | false | We calculate how many possible happy substrings of given length minus one exist \u2013 using this number, `k`, a list of possible letters and the last used letter we easily calculate what would be the letter at the current index.\n\nThis solution has O(n) runtime complexity and O(1) space complexity (result not counted... | 5 | 0 | ['Combinatorics', 'Python', 'Python3'] | 3 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Javascript Solution construct possible strings, sort and return kth string | javascript-solution-construct-possible-s-voxf | \n//contructs an array of all possible unique strings possible with \'a\', \'b\', \'c\' of length n\nvar formArray = function(res, s, pre, l, n){\n if (n == | seriously_ridhi | NORMAL | 2020-04-18T18:21:26.740492+00:00 | 2020-04-18T19:25:02.571929+00:00 | 311 | false | ```\n//contructs an array of all possible unique strings possible with \'a\', \'b\', \'c\' of length n\nvar formArray = function(res, s, pre, l, n){\n if (n == 0) \n { \n res.push(pre);\n return; \n } \n for (var i = 0; i < l; i++) \n { \n var newPre; \n \n // Next ch... | 5 | 1 | ['JavaScript'] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Easy python recursive + [sorting(lexicographically) / heap] | easy-python-recursive-sortinglexicograph-qaq4 | Approach 1 . Generating all the strings of length n (252 ms).\n\nclass Solution:\n def getHappyString(self, n: int, k: int) -> str:\n res=[] # lis | codes_in_dark | NORMAL | 2020-04-18T16:09:40.216295+00:00 | 2020-04-19T01:00:00.317102+00:00 | 518 | false | Approach 1 . Generating all the strings of length n (252 ms).\n```\nclass Solution:\n def getHappyString(self, n: int, k: int) -> str:\n res=[] # list storing all the strings of size n \n def fun(s, curr, n): # recurisve function for finding all strings\n if n==0 : \n ... | 5 | 0 | [] | 2 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | O(3^n) | Space (Only recusive stack) | Backtracking | o3n-space-only-recusive-stack-backtracki-n6cv | IntuitionApproachBacktracking TechniqueWe use backtracking to generate all valid happy strings of length n.Explore all possibilities by recursively adding chara | Developer_Ashish_1234 | NORMAL | 2025-02-19T06:35:21.441797+00:00 | 2025-02-19T06:35:21.441797+00:00 | 68 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
Backtracking Technique
We use backtracking to generate all valid happy strings of length n.
Explore all possibilities by recursively adding characters 'a', 'b', and 'c' to... | 4 | 0 | ['C++'] | 1 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Beats 100% ✅Easy C++ solution using backtracking | beats-100-easy-c-solution-using-backtrac-jq8a | IntuitionTry to make all the good string of size n. Return the kth lexicographically shortest .It simply implies that we can generate all the good string in lex | shivangkat07 | NORMAL | 2025-02-19T04:58:28.037591+00:00 | 2025-02-19T04:58:28.037591+00:00 | 190 | false | 
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Try to make all the good string of size n. Return the kth lexicographically shortest .It simply impli... | 4 | 0 | ['C++'] | 2 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | One Linear! | one-linear-by-charnavoki-8i9b | null | charnavoki | NORMAL | 2025-02-19T01:19:34.537035+00:00 | 2025-02-19T01:19:34.537035+00:00 | 414 | false |
```javascript []
const getHappyString = f = (n, k, list = ['']) =>
n ? f(n - 1, k, list.flatMap(v => [...'abc'].map(c => !v.endsWith(c) ? v + c : '').filter($=>$))) : list[k - 1] || '';
``` | 4 | 0 | ['JavaScript'] | 1 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Simple Clean Code with Description and recursive tree diagram. | simple-clean-code-with-description-and-r-sdo6 | Intuition\nBased on the recursive tree below we want to push the values into vector v.\n\n\n# Approach\nWe are going to save the current string in op and iterat | SSS009 | NORMAL | 2023-01-11T21:47:21.715542+00:00 | 2023-01-11T21:47:21.715579+00:00 | 472 | false | # Intuition\nBased on the recursive tree below we want to push the values into vector v.\n\n\n# Approach\nWe are going to save the current string in op and iterate over and over until the cur... | 4 | 0 | ['Backtracking', 'Recursion', 'C++'] | 1 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | C++, C# backtracking, dfs | c-c-backtracking-dfs-by-leonenko-5r6b | C++\n\nstring result;\nstring getHappyString(int n, int k) {\n\tstd::vector<char> str;\n\tdfs(str, n, k);\n\treturn result;\n}\n\nvoid dfs(std::vector<char>& st | leonenko | NORMAL | 2020-04-18T16:20:04.121388+00:00 | 2020-04-18T16:54:50.900991+00:00 | 570 | false | C++\n```\nstring result;\nstring getHappyString(int n, int k) {\n\tstd::vector<char> str;\n\tdfs(str, n, k);\n\treturn result;\n}\n\nvoid dfs(std::vector<char>& str, int n, int& k) {\n\tif (n == 0) {\n\t\tif (--k == 0) result = string(str.begin(), str.end());\n\t} else {\n\t\tfor(char ch = \'a\'; ch <= \'c\'; ++ch) {\n... | 4 | 1 | ['Backtracking', 'Depth-First Search', 'C', 'C++'] | 1 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | [Java] DP (recursive) + Heap | java-dp-recursive-heap-by-manrajsingh007-xe4v | ```\nclass Solution {\n public static List func(int n, int i, int prev, HashMap> map) {\n if(i == n) {\n List ans = new ArrayList<>();\n | manrajsingh007 | NORMAL | 2020-04-18T16:01:52.792216+00:00 | 2020-04-18T16:01:52.792268+00:00 | 250 | false | ```\nclass Solution {\n public static List<String> func(int n, int i, int prev, HashMap<Integer, List<String>> map) {\n if(i == n) {\n List<String> ans = new ArrayList<>();\n ans.add("");\n return ans;\n }\n if(map.containsKey(i * 10 + prev)) return map.get(i * 1... | 4 | 3 | [] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Optimized Lexicographic Happy String Generation in O(n) Time | optimized-lexicographic-happy-string-gen-ywnq | IntuitionThe problem requires generating a lexicographically ordered "happy string" of length n, where adjacent characters are different. Since there are only t | lanireddyhiteshreddy | NORMAL | 2025-02-19T12:22:24.665853+00:00 | 2025-02-19T12:22:24.665853+00:00 | 15 | false | # Intuition
The problem requires generating a lexicographically ordered "happy string" of length n, where adjacent characters are different. Since there are only three choices (a, b, c) at each position, we can generate all possible happy strings in lexicographic order and return the k-th one. Instead of generating all... | 3 | 0 | ['C++'] | 2 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | O(n) approach 100% beats, without recursion, using P&C to generate string | on-approach-100-beats-without-recursion-8e3ea | IntuitionFirst thought was only to generate all the possible strings and get to the answer, then I thought of optmization where i need to store all but just kee | v2khan | NORMAL | 2025-02-19T11:49:34.360346+00:00 | 2025-02-19T11:50:03.378043+00:00 | 78 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
First thought was only to generate all the possible strings and get to the answer, then I thought of optmization where i need to store all but just keep the counter on k to let it reach 0 and boom we got the string, but both these approache... | 3 | 0 | ['Combinatorics', 'C++'] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | ✅ Easy to Understand | Backtracking | Recursive Thinking | Detailed Video explanation🔥 | easy-to-understand-backtracking-recursiv-vccv | IntuitionThe problem requires generating lexicographically sorted "happy" strings of lengthnusing characters 'a', 'b', and 'c', where no two adjacent characters | sahilpcs | NORMAL | 2025-02-19T09:25:38.492748+00:00 | 2025-02-19T09:25:38.492748+00:00 | 160 | false | # Intuition
The problem requires generating lexicographically sorted "happy" strings of length `n` using characters 'a', 'b', and 'c', where no two adjacent characters are the same. The k-th happy string needs to be determined efficiently.
# Approach
We use a recursive backtracking approach to generate all possible ha... | 3 | 0 | ['String', 'Backtracking', 'Java'] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Generate all possible Happy Strings | generate-all-possible-happy-strings-by-d-qdc0 | How it works:We generate all possible combinations of length nReturn k - 1 possition or empty stringComplexity
Time complexity: O(3 ^ n) - In worst case
Space | Danil_M | NORMAL | 2025-02-19T08:46:33.096151+00:00 | 2025-02-19T08:46:33.096151+00:00 | 11 | false | **How it works:**
We generate all possible combinations of length n
Return k - 1 possition or empty string
# Complexity
- Time complexity: O(3 ^ n) - In worst case
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(n)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```kotlin [... | 3 | 0 | ['Kotlin'] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Math Approach || Clean code || O(n) RT and MEM | math-approach-clean-code-on-rt-and-mem-b-uo27 | Thanks to_Marat_for approach and python3-codeIntuitionYou can count the total number of happy strings depent by size (n).totalCount=3∗2n−1Because all of 3 chars | Rixaki | NORMAL | 2025-02-19T08:33:26.174982+00:00 | 2025-02-19T12:13:52.001185+00:00 | 242 | false | Thanks to_ [Marat](https://leetcode.com/u/iiByEb5iOA/) _for approach and python3-code
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
You can count the total number of happy strings depent by size (n).
$$
totalCount = 3*2^{n-1}
$$
Because all of 3 chars avaliable for first place ... | 3 | 0 | ['Math', 'String', 'Java', 'Python3', 'Kotlin'] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Simple and Easy Approach by using Recursion (Easy to Understand) | simple-and-easy-approach-by-using-recurs-ynic | IntuitionThe problem requires generating "happy strings" of length n using the characters {a, b, c} such that no two consecutive characters are the same. We nee | hariomsharma70 | NORMAL | 2025-02-19T08:18:23.038967+00:00 | 2025-02-19T08:18:23.038967+00:00 | 7 | false | # Intuition
The problem requires generating "happy strings" of length n using the characters {a, b, c} such that no two consecutive characters are the same. We need to return the k-th lexicographically smallest happy string.
A brute-force approach would be generating all possible valid strings and then returning the k... | 3 | 0 | ['Java'] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Backtracking approach. Beats 100% | backtracking-approach-beats-100-by-nnzl4-xyhu | IntuitionThe problem requires generating lexicographically sorted happy strings of lengthnand returning thek-th string. Since happy strings are built with const | nnzl48NW9N | NORMAL | 2025-02-19T05:29:31.263094+00:00 | 2025-02-19T05:29:31.263094+00:00 | 57 | false | # Intuition
The problem requires generating lexicographically sorted happy strings of length `n` and returning the `k`-th string. Since happy strings are built with constraints, a backtracking approach is suitable for generating valid strings while counting them.
Space complexity:O(3n)
⬆️👇⬆️UPVOTE it⬆️👇⬆️Code⬆️👇⬆️UPVOTE it⬆️👇⬆️ | shobhit_yadav | NORMAL | 2025-02-19T04:00:45.767872+00:00 | 2025-02-19T04:00:45.767872+00:00 | 184 | false | # Complexity
- Time complexity: $$O(3^n)$$
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: $$O(3^n)$$
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
⬆️👇⬆️UPVOTE it⬆️👇⬆️
# Code
```cpp []
class Solution {
public:
string getHappyString(int n, int k) {
const unordered_map<char, ... | 3 | 0 | ['String', 'Backtracking', 'C++', 'Python3'] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Simple & Easy C++ Solution || Backtracking | simple-easy-c-solution-backtracking-by-s-kuwn | if it's help, please up ⬆ vote! ❤️Code | shishirRsiam | NORMAL | 2025-02-19T02:52:05.853405+00:00 | 2025-02-19T02:52:05.853405+00:00 | 313 | false |
# if it's help, please up ⬆ vote! ❤️
# Code
```cpp []
class Solution {
public:
string s = "abc", ans;
void backtrack(string cur, int &n, int &k, int &count)
{
if(count == k) return;
if(cur.size() == n)
{
ans = cur, count += 1;
return;
}
for(... | 3 | 0 | ['String', 'Backtracking', 'Greedy', 'Recursion', 'C++'] | 2 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Basic loop and Most easiest and intuitve solution you can find. Beats 100% | basic-loop-and-most-easiest-and-intuitve-bh4b | IntuitionMy basic intuition is based on simple number of combinations. By simple law of counting(Permuatations), the total number of strings possible are 3*(2^( | avinoor19460 | NORMAL | 2025-02-19T01:00:13.511333+00:00 | 2025-02-19T01:00:13.511333+00:00 | 106 | false | # Intuition
My basic intuition is based on simple number of combinations. By simple law of counting(Permuatations), the total number of strings possible are 3*(2^(n - 1)) where n is length of string. Now if we fix one character position at a time we can determine number of permutations from there on, and can subtract t... | 3 | 0 | ['C++'] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Binary Partitioning Trick to Find the k-th Lexicographical Happy String in O(n) Time | binary-partitioning-trick-to-find-the-k-30h9w | IntuitionThe problem requires generating a lexicographically sorted list of "happy" strings of lengthnand returning thek-th one.A happy string is a string where | alperensumeroglu | NORMAL | 2025-02-19T00:50:34.404536+00:00 | 2025-02-19T00:50:34.404536+00:00 | 7 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem requires generating a lexicographically sorted list of "happy" strings of length `n` and returning the `k`-th one.
A happy string is a string where no two adjacent characters are the same and only consists of 'a', 'b', and 'c'.
... | 0 | 0 | ['Python3'] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | TOO MUCH AURA OVERLOAD!!! Beats 99% Java DFS Solution ULTIMATE RIZZZZZZZZZZZZZZZZZZZZZZZZZZZZ | beats-99-java-dfs-solution-by-ishaan_p-5pd9 | Just create the strings in lexicographic order using DFS, so we can avoid the sorting step altogether. Stop when we reach the kth string.Code | Ishaan_P | NORMAL | 2025-02-19T00:44:56.365073+00:00 | 2025-02-19T04:31:20.798338+00:00 | 582 | false | Just create the strings in lexicographic order using DFS, so we can avoid the sorting step altogether. Stop when we reach the kth string.
# Code
```java []
class Solution {
String result= "";
int count = 0;
public String getHappyString(int n, int k) {
buildHappyStrings(n,k,new StringBuilder());
... | 3 | 0 | ['Java'] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | C# - Fast - 0ms - binary/trinary radix math | c-fast-0ms-binarytrinary-radix-math-by-h-220s | IntuitionThis is mostly just a conversion of a number to binary since most (all but the first) letter can only be one of two values since the previous letter ma | hafthor | NORMAL | 2025-02-19T00:29:42.721311+00:00 | 2025-02-19T16:07:02.257617+00:00 | 247 | false | # Intuition
This is mostly just a conversion of a number to binary since most (all but the first) letter can only be one of two values since the previous letter makes one choice of three forbidden. If the bit is 0, we pick the lowest available letter. If the bit is 1, we pick the highest available letter. The first let... | 3 | 0 | ['Math', 'C#'] | 2 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | 🚀 Easy | Backtracking | JavaScript Solution | easy-backtracking-javascript-solution-by-zquc | Approach / Intuition
Backtracking Approach: We generate all possible happy strings of lengthnusing backtracking, ensuring that no two adjacent characters are th | dharmaraj_rathinavel | NORMAL | 2025-02-19T00:29:18.375873+00:00 | 2025-02-20T03:05:00.627532+00:00 | 106 | false | ## **Approach / Intuition**
- **Backtracking Approach**: We generate all possible happy strings of length `n` using backtracking, ensuring that no two adjacent characters are the same.
- **Recursive Exploration**: At each step, we append characters `'a'`, `'b'`, or `'c'` to the current string, skipping characters ... | 3 | 0 | ['String', 'Backtracking', 'JavaScript'] | 2 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Simple, intuitive solution with no backtracking in C++. | simple-intuitive-solution-with-no-backtr-h3xa | Approach\nWe have just mapped the string "abc" and traverse through every position so that to place appropriate letter at given word.\n\nWe can observe the patt | k_patil092 | NORMAL | 2023-08-03T04:42:24.887401+00:00 | 2023-08-03T04:42:24.887436+00:00 | 136 | false | # Approach\nWe have just mapped the string "abc" and traverse through every position so that to place appropriate letter at given word.\n\nWe can observe the pattern we will get after arranging in lexicographically order. For example n = 4:\n\n<"abab", "abac", "abca", "abcb">, <"acab", "acac", "acba", "acbc">,\n<"baba... | 3 | 0 | ['String', 'Brainteaser', 'Ordered Map', 'Suffix Array', 'Combinatorics', 'Interactive', 'Hash Function', 'C++'] | 2 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | ✅C++ || Same as Cobination sum-1 || Backtracking | c-same-as-cobination-sum-1-backtracking-e07ux | \n\nThis Que is same as Leetcode Combination sum-1\n\nT->O(2^n) && S->O(n) [Recursion Stack Space]\n\n\tclass Solution {\n\tpublic:\n\t\tvector v;\n\t\tvoid f(i | abhinav_0107 | NORMAL | 2022-08-20T06:33:26.928054+00:00 | 2022-08-20T06:33:26.928103+00:00 | 460 | false | \n\n***This Que is same as Leetcode Combination sum-1***\n\n**T->O(2^n) && S->O(n) [Recursion Stack Space]**\n\n\tclass Solution {\n\tpublic:\n\t\tvector<string> v;\n\t\tvoid f(int i,string& s,int n){\n\t\t\tif... | 3 | 0 | ['Backtracking', 'Recursion', 'C', 'C++'] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | [Python] Bruteforce and optimized solution | python-bruteforce-and-optimized-solution-occn | Brute force solution by generating all possible combinations and selecting the kth entry. O(2^n) [580ms]\n\nclass Solution(object):\n def getHappyString(self | johnnydu | NORMAL | 2020-04-20T01:17:19.872409+00:00 | 2020-04-20T01:23:32.065255+00:00 | 278 | false | **Brute force solution by generating all possible combinations and selecting the kth entry. O(2^n) [580ms]**\n```\nclass Solution(object):\n def getHappyString(self, n, k):\n """\n :type n: int\n :type k: int\n :rtype: str\n """\n result = self.listCombinations(n, \'\')\n ... | 3 | 0 | [] | 2 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | [Java] Simple backtracking | java-simple-backtracking-by-rajeshtomar-vfkh | Just backtrack to generate all the strings and store them in a list. Return the appropriate element from list.\n\nclass Solution {\n \n List<String> list | rajeshtomar | NORMAL | 2020-04-18T17:01:08.591599+00:00 | 2020-04-18T17:01:08.591652+00:00 | 428 | false | Just backtrack to generate all the strings and store them in a list. Return the appropriate element from list.\n```\nclass Solution {\n \n List<String> list = new ArrayList<>();\n \n void rec(int n, StringBuilder sb) {\n if (n == 0) {\n list.add(sb.toString());\n return;\n ... | 3 | 0 | [] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Java O(n) solution | java-on-solution-by-s_tsat-jida | The solution is based on the fact that there are 3 choices for the first character and 2 for each subsequent one.\nOn each step, I append the corresponding char | s_tsat | NORMAL | 2020-04-18T16:31:30.996011+00:00 | 2020-06-29T22:29:50.400933+00:00 | 338 | false | The solution is based on the fact that there are 3 choices for the first character and 2 for each subsequent one.\nOn each step, I append the corresponding character based on k.\n\n```\n public String getHappyString(int n, int k) {\n \n k--; // this is to make arithmetic a bit easier\n int powe... | 3 | 0 | ['Java'] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | [Python] Fast solution using divmod - No DFS (100% / 100%) | python-fast-solution-using-divmod-no-dfs-8wuu | The idea of this solution comes from 60. Permutation Sequence. \n\n\nclass Solution:\n def getHappyString(self, n: int, k: int) -> str:\n # now k star | alanlzl | NORMAL | 2020-04-18T16:21:43.147779+00:00 | 2020-04-30T04:11:48.530872+00:00 | 283 | false | The idea of this solution comes from [60. Permutation Sequence](https://leetcode.com/problems/permutation-sequence/). \n\n```\nclass Solution:\n def getHappyString(self, n: int, k: int) -> str:\n # now k starting from index 0\n k -= 1\n \n bucket_size = (2 ** (n - 1))\n bucket_inde... | 3 | 0 | [] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | [C++] Simple search for k-th string O(n) | c-simple-search-for-k-th-string-on-by-ni-palo | Here we always keep a count of strings starting with a, b, and c .i.e. ca, cb, cc respectively.\nWe maintain these counts and for every index we see what charac | nicmit | NORMAL | 2020-04-18T16:02:52.221470+00:00 | 2020-04-18T16:04:25.227195+00:00 | 293 | false | Here we always keep a count of strings starting with a, b, and c .i.e. `ca, cb, cc` respectively.\nWe maintain these counts and for every index we see what character needs to come here. This is similar to dictionary idea.\n```\nclass Solution {\npublic:\n int f(int n)\n {\n if(n == 0)\n return 0... | 3 | 1 | [] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Beats 100% || step by step clearly Explained | beats-100-clearly-explained-by-aadithya1-ep7t | IntuitionKnow which character will be our next character and simply add it.ApproachWe have only 3 characters we can use.So, the valid words of n = 1 will be:a,b | aadithya18 | NORMAL | 2025-02-19T17:53:22.951120+00:00 | 2025-02-19T20:46:46.168855+00:00 | 9 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Know which character will be our next character and simply add it.
# Approach
<!-- Describe your approach to solving the problem. -->
We have only 3 characters we can use.
So, the valid words of n = 1 will be: `a,b,c`. in the same order.
F... | 2 | 0 | ['String', 'Backtracking', 'C++'] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Combinatorics Trick [Python] | Obligatory Beats 100% | combinatorics-trick-by-shirsho12-ctgr | IntuitionNotice that in this question, we have 3 choices for the first character, and 2 for each of the subsequent ones. In fact, as the result is ordered, we n | shirsho12 | NORMAL | 2025-02-19T17:33:14.639725+00:00 | 2025-02-19T17:35:26.285712+00:00 | 16 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Notice that in this question, we have 3 choices for the first character, and 2 for each of the subsequent ones. In fact, as the result is ordered, we notice that `0` -> smaller character, `1` -> larger character.
# Approach
<!-- Describe... | 2 | 0 | ['String', 'Combinatorics', 'Python3'] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | beats 100% || very simple logic | beats-100-very-simple-logic-by-shashwat1-d2rd | Code | shashwat1915 | NORMAL | 2025-02-19T16:18:24.332408+00:00 | 2025-02-19T16:18:24.332408+00:00 | 23 | false | # Code
```cpp []
class Solution {
public:
string ans="";
void get_ans(int n,int &k,string &ans1,vector<char>&arr){
if(!ans.empty()) return;
if(ans1.length()>n) return;
if(k==0 && ans1.length()==n){
ans=ans1;
return;
}
int x=-1;
if(!ans1.empty()){
if(ans1.back()=='a') x=0;
if(ans1.back()=... | 2 | 0 | ['String', 'Backtracking', 'C++'] | 0 |
the-k-th-lexicographical-string-of-all-happy-strings-of-length-n | Easy Solution || Simple Approach || C++ || Backtracking | easy-solution-simple-approach-c-backtrac-u256 | IntuitionWe need to generate all "happy strings" of length n, where a happy string is one that does not have two identical consecutive characters. Instead of tr | Santhosh_TechWriter | NORMAL | 2025-02-19T15:19:27.428599+00:00 | 2025-02-19T15:19:27.428599+00:00 | 29 | false | # Intuition
We need to generate all "happy strings" of length n, where a happy string is one that does not have two identical consecutive characters. Instead of trying to generate all strings and then filtering out the invalid ones, we can build each string character by character while ensuring we never add the same ch... | 2 | 0 | ['Backtracking', 'C++'] | 0 |
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